Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Answer:
1. True
2. False
3. False
4. True
Explanation:
A lab is an abbreviation for laboratory and it can be defined as a building that is equipped with scientific equipments and apparatus, which are used for performing scientific experiments, manufacturing of chemicals and drugs, research, testing and analysis of hypothesis in order to establish a theory.
Generally, laboratory scientists, technicians, students and anyone going into a laboratory is prohibited from taking in edible items such as food and water.
Basically, all laboratories are required to place a visible warning sign around the lab prohibiting the bringing in and ingestion of food in any form.
1. True: Tobacco products of all kinds are prohibited in lab.
2. False: Beverages are allowed in lab as long as they are in containers with lids. It's completely prohibited from the lab so as to prevent food poisoning.
3. False: Chewing gum is okay in the lab as long as you don't blow bubbles.
4. True: Food items should not be brought into the lab.
The following is produced when propane (C₃H₈) is combusted completely : H₂O
<h3>Further explanation </h3>
Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (especially alkanes) 
For combustion of propane C₃H₈ (n = 3) ⇒ completely(excess O₂) :
C₃H₈+5O₂⇒3CO₂+4H₂O
The products of combustion : CO₂ and H₂O
Anesthetic in dentistry consists of a mixture of dinitrogen oxide (N₂O) and oxygen gas (O₂), which is administered through an inhaler over the nose. Total pressure of the mixture (
) is sum of partial pressure of N₂O (
) and partial pressure of O₂, (
).
=370 mmHg
So, = (740-370) mmHg= 370 mmHg=370 torr
Hence, partial pressure of N₂O is 370 torr.
As, 1mmHg= 1 torr.
We can express the rate equation in this form:
-r = k A^n B^m
where -r is the rate
k is the rate constant,
A is the concentration of CH3Cl
n is the order with respect to CH3Cl
B is the concentration of H2O
m is the order with respect to H2O
We can solve this by trial and error or by calculus. The first method is easier. The rate constant does not depend on the concentration of the reactant. Assume values of n and m and solve for k in each experiment. The only option that gives really close values of k in each experiment is:
<span>C. CH3Cl: firstorder H2O: second order
</span>
